Problem: Compute the domain of the real-valued function $$f(x)=\sqrt{3-\sqrt{5-\sqrt{x}}}.$$
Solution: For the contents of the innermost square root to be nonnegative, we must have $x\geq 0$.  To satisfy the middle square root, we must have  $$5-\sqrt{x}\geq 0$$ $$\Rightarrow 25\geq x.$$ Finally, the outermost square root requires $$3-\sqrt{5-\sqrt{x}}\geq 0$$ or $$9\geq 5-\sqrt{x}$$ $$\Rightarrow \sqrt{x}\geq -4,$$ which is always true. Combining our inequalities, we get $$0\leq x\leq 25,$$ or $x \in \boxed{[0, 25]}$ in interval notation.